z tangensa na cosinus

mol_ksiazkowy · Post autor: **mol_ksiazkowy** » 3 gru 2006, o 13:36

Oblicz \(\displaystyle{ \frac{22}{4+39cos(2\alpha)}}\), jesli \(\displaystyle{ tg(\alpha)=\frac{3}{2}}\)

Lorek · Post autor: **Lorek** » 3 gru 2006, o 14:00

\(\displaystyle{ \cos 2\alpha=\cos^2\alpha-\sin^2\alpha=\cos^2\alpha-tg^2\alpha\cos^2\alpha=\cos^2\alpha(1-tg^2\alpha)=\\\\=\frac{1-tg^2\alpha}{\frac{1}{\cos^2\alpha}}=\frac{1-tg^2\alpha}{\frac{\sin^2\alpha+\cos^2\alpha}{\cos^2\alpha}}=\frac{1-tg^2\alpha}{1+tg^2\alpha}\\\\\cos 2\alpha=\frac{1-\frac{9}{4}}{1+\frac{9}{4}}=-\frac{5}{13}\\\\\frac{22}{4+39\cos 2\alpha}=\frac{22}{4+39(\frac{-5}{13})}=\frac{22}{-11}=-2}\)